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So the question is really hard I think. I tried using a simple way by calculating the probability of each combination that makes a sum divisible by six, but it would take forever. Does anyone have any ideas?

Suppose that we roll a six-sided die ten times. What is the probability that the total of all ten rolls is divisible by six?

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  • $\begingroup$ Almost a duplicate of this question. Using the same logic, the answer is easily seen to be $1/6$. $\endgroup$
    – TonyK
    Commented Nov 11, 2015 at 15:33

7 Answers 7

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Hint.

Roll $9$ times and let $x$ be the total.

For exactly one number $n\in\{1,2,3,4,5,6\}$ we will have $6 \mid (x+n)$ (i.e. $x+n$ is divisible by $6$).

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    $\begingroup$ I didn't get it, why would I roll 9 times instead of 10? and what is 6|x? $\endgroup$
    – Xlyon
    Commented Nov 11, 2015 at 9:38
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    $\begingroup$ @Xlyon When you've rolled nine times there's always exactly one outcome for the tenth that will make the sum divisible by $6$. $\endgroup$
    – skyking
    Commented Nov 11, 2015 at 9:43
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    $\begingroup$ Thanks! that is logical. Can you please help me complete it? $\endgroup$
    – Xlyon
    Commented Nov 11, 2015 at 10:50
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    $\begingroup$ Exactly one of the faces of the die thrown at the $10$th time will result in a total sum ($x+n$) that is divisible by $6$. The probability that this face shows up is $\frac16$ (if the die is fair, of course). $\endgroup$
    – drhab
    Commented Nov 11, 2015 at 10:54
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    $\begingroup$ The value of $x$ is indeed irrelevant. Whatever value $x$ takes, it's chance to change into a number divisible by $6$ (when the last result is added by $x$) is in all cases $\frac16$. So that's indeed the answer to this question. They could have asked: take some arbitrary number $y\in\mathbb Z$ and trhrow a fair die. If $D$ denotes the outcome then what is the probability that $y+D$ is divisible by $6$? Same answer: $\frac16$, $\endgroup$
    – drhab
    Commented Nov 11, 2015 at 11:34
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After rolling the die once, there is equal probability for each result modulo 6. Adding any unrelated integer to it will preserve the equidistribution. So you can even roll a 20-sided die afterwards and add its outcome: the total sum will still have a probability of 1/6 to be divisible by 6.

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    $\begingroup$ @Kevin The d20 by itself does have that unequal distribution, but the sum d20+d6 does not. It literally does not matter what _**x**_ is in determining the distribution of (_**x**_ + d6) mod 6. In fact, d20 + d6 also has exactly a 1/20 chance of being divisible by 20 by the same logic $\endgroup$ Commented Nov 11, 2015 at 19:28
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If you want something a little more formal and solid than drhab's clever and brilliant answer:

Let $P(k,n)$ be the probability of rolling a total with remainder $k$ when divided by $6, (k = 0...5)$ with $n$ die.

$P(k, 1)$ = Probability of rolling a $k$ if $k \ne 0$ or a $6$ if $k = 6$; $P(k, 1) = \frac 1 6$.

For $n > 1$. $ P(k,n) = \sum_{k= 0}^5 P(k, n-1)\cdot \text{Probability of Rolling(6-k)} = \sum_{k= 0}^5 P(k, n-1)\cdot\frac 1 6= \frac 1 6\sum_{k= 0}^5 P(k, n-1)= \frac 1 6 \cdot 1 = \frac 1 6$

This is drhab's answer but in formal terms without appeals to common sense

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    $\begingroup$ That's a way to do it. What I actually had in mind was: $P\left(6\text{ divides }S_{10}\right)=\sum_{m\in\mathbb{N}}P\left(6\text{ divides }S_{10}\mid S_{9}=m\right)P\left(S_{9}=m\right)=\sum_{m\in\mathbb{N}}\frac{1}{6}P\left(S_{9}=m\right)=\frac{1}{6}$ $\endgroup$
    – drhab
    Commented Nov 12, 2015 at 9:40
  • $\begingroup$ Yes, that is more direct. $\endgroup$
    – fleablood
    Commented Nov 12, 2015 at 16:45
  • $\begingroup$ This should be the accepted answer. It's formulated in accessible notation and actually makes sense. I did not trust the "common sense" argument in the answer of @drhab (and their comment to this answer is opaque to me). $\endgroup$
    – Kagaratsch
    Commented Mar 1, 2021 at 21:00
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    $\begingroup$ I dunno. If you can simplify something done to an easy, almost trivial case, you should. A $\sum_{k=1}^{10} a_k \equiv 0 \pmod 6 \iff \sum_{k=1}^{9}a_k \equiv - a_{10}\pmod 6$. And for any value $M = \sum_{k=1}^{9}a_k$ we know the probablilty that $a_{10}\equiv M \pmod 6$ is exactly $\frac 16$. ... Now if I teaching probability I don't want students shying away from rolling up the sleeves and getting dirty but but If I want to do an answer I want things as simple as possible. If it doesn't matter what the first nine rolls are so long as the 10th cancels them... then why bother. $\endgroup$
    – fleablood
    Commented Mar 1, 2021 at 21:35
  • $\begingroup$ @Kagaratsch Drhab has written down his steps in rigorous and standard mathematical language. Why didn't you trust? On the contrast, I can't really follow the notations here and why it is true that $P(k,n) = \sum_{k= 0}^5 P(k, n-1)\cdot \text{Probability of Rolling(6-k)} $. $\endgroup$
    – Sam Wong
    Commented May 28 at 9:29
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In spite of all great answers, given here, I say, why not give another proof, from another point of view. The problem is we have 10 random variables $X_i$ for $i=1,\dots,10$, defined over $[6]=\{1,\dots,6\}$, and we are interested in distribution of $Z$ defined as $$ Z=X_1\oplus X_2\oplus \dots \oplus X_{10} $$ where $\oplus$ is addition modulo $6$. We can go on by two different, yet similar proofs.


First proof: If $X_1$ and $X_2$ are two random variables over $[6]$, and $X_1$ is uniformly distributed, sheer calculation can show that $X_1\oplus X_2$ is also uniformly distributed. Same logic yields that $Z$ is uniformly distributed over $[6]$.

Remark: This proves a more general problem. It says that even if only one of the dices is fair dice, i.e. each side appearing with probability $\frac 16$, the distribution of $Z$ will be uniform and hence $\mathbb P(Z=0)=\frac 16$.


Second proof: This proof draws on (simple) information theoretic tools and assumes its background. The random variable $Z$ is output of an additive noisy channel and it is known that the worst case is uniformly distributed noise. In other word if $X_i$ is uniform for only one $i$, $Z$ will be uniform. To see this, suppose that $X_1$ is uniformly distributed. Then consider the following mutual information $I(X_2,X_3,\dots,X_6;Z)$ which can be written as $H(Z)-H(Z|X_2,\dots,X_6)$. But we have: $$ H(Z|X_2,\dots,X_6)=H(X_1|X_2,\dots,X_6)=H(X_1) $$
where the first equality is resulted from the fact that knowing $X_2,\dots,X_6$ the only uncertainty in $Z$ is due to $X_1$. The second equality is because $X_1$ is independent of others. Know see that:

  • Mutual information is positive: $H(Z)\geq H(X_1)$
  • Entropy of $Z$ is always less that or equal to the entropy of uniformly distributed random variable over $[6]$: $H(Z)\leq H(X_1)$
  • From the last two $H(Z)=H(X_1)$ and $Z$ is uniformly distributed and the proof is complete.

Similarly here, only one fair dice is enough. Moreover the same proof can be used for an arbitrary set $[n]$. As long as one of the $X_i$'s is uniform, then their finite sum modulo $n$ will be uniformly distributed.

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There are 3 variables in this case:

  • the number of sides of the dice: s (e.g. 6)
  • the number of throws: t (e.g. 10)
  • the requesed multiple: x (e.g. 6)

In this case, the conditions are simple:

  • s>=x
  • x >0
  • t > 0

And also the answer is simple: Throwing a sum that is a multiple of 6 has a 1/6 probability.

$P(s,t,x) = 1/x$

For situations where s<x this is not entirely correct. It approaches the same result though, at a high amount of throws. Example: If you throw a 6-sided dice 30 times the chance that the sum is a multiple of 20 will be about 5%. Proving this is a bit of a challenge.

$\lim \limits_{t \to \infty} P(s,t,x) = 1/x$,

Nevertheless, if programming is an acceptable proof:

public static void main(String[] args) {
    int t_throws = 10;
    int s_sides = 6;
    int x_multiple = 6;
    int[] diceCurrentValues = new int[t_throws];
    for (int i = 0; i < diceCurrentValues.length; i++) diceCurrentValues[i] = 1;

    int combinations = 0;
    int matches = 0;
    for (; ; ) {
        // calculate the sum of the current combination
        int sum = 0;
        for (int diceValue : diceCurrentValues) sum += diceValue;

        combinations++;
        if (sum % x_multiple == 0) matches++;
        System.out.println("status: " + matches + "/" + combinations + "=" + (matches * 100 / (double) combinations) + "%");

        // create the next dice combination
        int dicePointer = 0;
        boolean incremented = false;
        while (!incremented) {
            if (dicePointer == diceCurrentValues.length) return;
            if (diceCurrentValues[dicePointer] == s_sides) {
                diceCurrentValues[dicePointer] = 1;
                dicePointer++;
            } else {
                diceCurrentValues[dicePointer]++;
                incremented = true;
            }
        }
    }
}

EDIT:

Here's another example. If you throw a 6-sided dice 10 times, there is 1/4 probability that the sum is a multiple of 4. The program above should run with the following parameters:

    int t_throws = 10;
    int s_sides = 6;
    int x_multiple = 4;

The program will show the final output: status: 15116544/60466176=25.0% That means that there are 60466176 combinations (i.e. 6^10) and that there are 15116544 of them where the sum is a multiple of 4. So, that's 25% (=1/4).

This just follows the formula as mentioned above (i.e. P(s,t,x) = 1/x). x is 4 in this case.

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  • $\begingroup$ "Assuming that t converges to infinity. This is also the case when s<x." This sounds nonsensical. $\endgroup$
    – djechlin
    Commented Nov 11, 2015 at 17:43
  • $\begingroup$ Excuse me for my poor English :) The thing is, if you apply this for values greater than the number of sides of your dice. (e.g. multiples of 10 with a 6 sided dice.) then P = 1/x is no longer correct. But it does converge 1/x ; meaning that if you would throw an infinit amount of times, the result would be 1/x again. $\endgroup$
    – bvdb
    Commented Nov 12, 2015 at 0:10
  • $\begingroup$ I made some slight adjustments. Let me know what you think. $\endgroup$
    – bvdb
    Commented Nov 12, 2015 at 0:22
  • $\begingroup$ It's not the case that the limiting factor is if s>=x. Consider the probability that 10 rolls of a six-sided dice divides 4. (Of course the limit still converges when the number of dices increases) $\endgroup$
    – Taemyr
    Commented Nov 12, 2015 at 8:40
  • $\begingroup$ @Taemyr for 10 roles with a 6-sided dice, there are 60466176 combinations, of which 15116544 have a sum which is a multiple of 4. That's exactly 25% which is exactly 1/4. So, it's correct, right ? $\endgroup$
    – bvdb
    Commented Nov 14, 2015 at 15:51
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Roll the die 9 times and add up the dots. The answer is x. Roll the die one more time. add the number thrown to x to get one and only one of the following answers; x+1, x+2, x+3, x+4, x+5 or x+6. since these answers are six sequential numbers one and only one of them will be divisible by six. Therefore the probability of the sum of ten rolls of a die being divisible by six is exactly 1/6.

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Couldn't you also think of it as the maximum possible value you could get for rolling the die 10 times would be 60, how many numbers between 1 and 60 are divisible by 6? 10 numbers are divisible by 6(6*1, 2, 3, etc.) So, 10 out of 60 possible values gives... 1/6. I love math.

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    $\begingroup$ True, but some numbers appear multiple times. So, even though your answer is correct, your logic is flawed. $\endgroup$
    – bvdb
    Commented Nov 11, 2015 at 11:58
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    $\begingroup$ The minimum possible value is 10. So only 9 of 51 values are divisible by 6. But not all numbers are equally probable. $\endgroup$
    – Cephalopod
    Commented Nov 11, 2015 at 11:58
  • $\begingroup$ The numbers are not distributed evenly. $\endgroup$
    – fleablood
    Commented Nov 11, 2015 at 20:31

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